Binary tree of splitting that separates point over every set? Is the following true?
Let I be any set. For me a binary tree of splitting of I will be the following:
start with $I_0=I$, at the step $n+1$ take the set of step $n$ and split each of them in two parts of equal cardinality. I say that a binary tree of splitting separates points if given every two points $x,y \in I$ there is an $n$ positive integer such that $x,y$ belongs to different sets at the n-th splitting.
Is it true that every sets admit a binary tree of splitting that separates points?
p.s: just after posting i realized that probably there's a trivial upper bound namely the continuum for this set.
 A: If $I$ is "splitting", its size is at most the size of the continuum. 
Let $2^{< \omega}$ be the set of all finite $0,1$ sequences and let $(I_s \mid s \in 2^{<\omega})$ be a witnessing splitting sequence for $I$. By this I mean


*

*$I_{(\ )} = I$, where $( \ )$ denotes the empty $0,1$ sequence

*$I_s = I_{s \frown 0} \cup I_{s \frown 1}$, where $s \frown i$ denotes the sequence $s$ with an additional last entry $i$, where $i \in \{0,1\}$

*All the $I_s$ have the same cardinality

*For all $x \neq y$ in $I$, there is some $s \in 2^{< \omega}$ s.t. $x,y \in I_s$, $x \in I_{s \frown 0} \setminus I_{s \frown 1}$ and $y \in I_{s \frown 1} \setminus I_{s \frown 0}$ (note that we may have to exchange the role of $x$ and $y$). Denote this $s$ by $x \Delta y$.


(This is just a handy notation for the binary tree you have in mind.)
Now
$$
f \colon I \to \mathcal P (2^{< \omega}), \ x \mapsto \{s \in 2^{< \omega} \mid x \in I_s \}
$$ 
is an injection as for $x \neq y$ the sequence $x \Delta y \frown 0$ is in exactly one of the sets $f(x)$ or $f(y)$.
Noting that $2^{< \omega}$ is countable, the claim follows.
